# A Few Statics-Related Questions

1. Dec 3, 2005

### amcavoy

I have a few questions related to statics and rotational dynamics. I have the general idea of what I need to do (at least, I hope I do), but I am having trouble making anything of it. Any input on any of the questions would be greatly appreciated .

The first thing I did was to look at the hanging mass and determine the forces acting on it. I said that the net force is T-1600N=0. Now for the system to remain in equilibrium, I made some assumptions: First, I though of S as a vector. For that vector to remain still, the torque produced by the tension in the direction of the point where the cable is attached to the ground and the torque produced by the tension in the direction of the hanging mass (downwards) must be zero. However, using the first equation I wrote above, I come up with the tension being equal to 1600N. Am I on the right track here?

I know that the answer to part (c) is RIGHT because it is rolling right producing a contact force to the left, thus a frictional force to the right. This is how I attempted to calculate part (a):

$$\tau=rF=\left(.05\text{N}\right)\left(.01\text{m}\right)=.0005\,\text{Nm}$$

...now dividing by the larger radius I can find the contact force, which should be equal to the frictional force in magnitude (is this right?):

$$\frac{.0005\,\text{Nm}}{.026\,\text{m}}=.019\,\text{N}$$

Thank you very much for the help.

Last edited by a moderator: Apr 21, 2017
2. Dec 3, 2005

### lightgrav

1) They want to know the Tension in the cable with they label "T" on the diagram. It is NOT necessarily the same Tension as in the hanging cable.
You have to ensure that the sum of torques by these two Forces is zero
(around the pivot at the base). Be careful with the angles between r_vec
and the T_vec .

2)
Do you mean to say that the frictional Force applied to the yo-yo is th the right? So the friction Force pushes the bottom of the yo-yo to the right?
Wouldn't that make the yo-yo SLIDE instead of roll?

Where do you think the "axis of rotation" is for this yo-yo?
recall that the axis of rotation does not move with small angle changes.
the r_vector extends from the axis of rotation to the point that the Force is applied.

3. Dec 4, 2005

### amcavoy

OK, as for the rolling yo-yo: I see that it will roll to the right. I have come up with the following:

$$F-F_{S}=ma$$

$$Fr-F_{S}R=mRa$$

But from these I always have two unknows (FS and a). Any ideas on this problem?

Thanks again.

4. Dec 5, 2005

### Staff: Mentor

OK.

Check this one. The right side should equal $I \alpha$.

Two equations and two unknowns. What's the problem?